Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.
For example:
Given n = 5 and edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return true.
Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], return false.
Note: you can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.
避免回访前驱：
recursive方式：传进parent来判断是不是nextNeighbor != parent?
stack：push的时候也push parent
Solution1：DFS recursive 找global && 所有vertices相连
思路：DFS 看global上是否不存在cycle && 并且所有vertices都是相连的
区别于207. Course Schedule http://www.jianshu.com/p/8665248b4458 onpath上cycle
写法1_a: 递归内部置visited=true
写法1_b: 在递归前置visited=true
Time Complexity: O(E+V) Space Complexity: O(E+V)
Solution2：similar to solution3
Time Complexity: O(E+V) Space Complexity: O(E+V)
Solution3：DFS(stack) 看global上是否不存在cycle
All same with solution2, just replace queue with stack
思路：DFS 看global上是否不存在cycle(通过检查是否visited过，但要避免check前驱， push的时候也push parent，来判断是不是nextNeighbor != parent ）
3_b: 内部置visited=true
Time Complexity: O(E+V) Space Complexity: O(E+V)
Solution4：Union Find 看global上是否不存在cycle
思路： 根据给每个边，每个结点都assign不同的id，union相连的，如果碰到有相同的id，说明有cycle。因为只在给的边上做，没有边的isolated结点 检查可以用edges.length == n - 1判断是否存在单独isolated的结点。而solution1-3不可以用此判断，因为不一定检查到所有的边，只能检查某块edge，这样就可能出现先单独结点dfs/bfs成功，但是也满足edges.length == n - 1，但其实别的块有loop。
实现4b: 高票简版
Time Complexity: O(ElogV) ? Space Complexity: O(V) ?
Solution1a Code：
class Solution {
public boolean validTree(int n, int[][] edges) {
// init graph (adjList)
List> adjList = new ArrayList>(n);
for (int i = 0; i < n; i++)
adjList.add(i, new ArrayList());
// add edges
for (int i = 0; i < edges.length; i++) {
int u = edges[i][0], v = edges[i][1];
adjList.get(u).add(v);
adjList.get(v).add(u);
}
boolean[] visited = new boolean[n];
visited[0] = true;
// make sure there's no cycle
if (dfsCycle(adjList, 0, visited, -1)) {
return false;
}
// check if all nodes are connected
for (int i = 0; i < n; i++) {
if (!visited[i]) {
return false;
}
}
return true;
}
// dfs checking if there is cycle
boolean dfsCycle(List> adjList, int cur, boolean[] visited, int prev) {
visited[cur] = true;
for (int i = 0; i < adjList.get(cur).size(); i++) {
int neighbor = adjList.get(cur).get(i);
if (neighbor == prev) {
continue;
}
else if (visited[neighbor]) {
return true;
}
else {
if (dfsCycle(adjList, neighbor, visited, cur)) {
return true;
}
}
}
return false;
}
}
Solution1b Code：
class Solution {
public boolean validTree(int n, int[][] edges) {
// init graph (adjList)
List> adjList = new ArrayList>(n);
for (int i = 0; i < n; i++)
adjList.add(i, new ArrayList());
// add edges
for (int i = 0; i < edges.length; i++) {
int u = edges[i][0], v = edges[i][1];
adjList.get(u).add(v);
adjList.get(v).add(u);
}
boolean[] visited = new boolean[n];
visited[0] = true;
// make sure there's no cycle
if (dfsCycle(adjList, 0, visited, -1)) {
return false;
}
// check if all nodes are connected
for (int i = 0; i < n; i++) {
if (!visited[i]) {
return false;
}
}
return true;
}
// dfs checking if there is cycle
boolean dfsCycle(List> adjList, int cur, boolean[] visited, int prev) {
for (int i = 0; i < adjList.get(cur).size(); i++) {
int neighbor = adjList.get(cur).get(i);
if (neighbor == prev) {
continue;
}
else if (visited[neighbor]) {
return true;
}
else {
visited[neighbor] = true;
if (dfsCycle(adjList, neighbor, visited, cur)) {
return true;
}
}
}
return false;
}
}
Solution2 Code：
TODO
Solution3 Code：
class Solution {
private class NodeWithParent {
int id;
int parentId;
public NodeWithParent(int id, int parentId) {
this.id = id;
this.parentId = parentId;
}
}
public boolean validTree(int n, int[][] edges) {
// init graph (adjList)
List> adjList = new ArrayList>(n);
for (int i = 0; i < n; i++)
adjList.add(i, new ArrayList());
for (int i = 0; i < edges.length; i++) {
int u = edges[i][0], v = edges[i][1];
adjList.get(u).add(v);
adjList.get(v).add(u);
}
// dfs
boolean[] visited = new boolean[n];
Deque stack = new ArrayDeque();
stack.push(new NodeWithParent(0, -1));
visited[0] = true;
while (!stack.isEmpty()) {
NodeWithParent cur = stack.pop();
for (int neighborId: adjList.get(cur.id)) {
if (neighborId == cur.parentId) {
continue;
}
else if ((visited[neighborId])) {
return false;
}
else {
stack.push(new NodeWithParent(neighborId, cur.id));
visited[neighborId] = true;
}
}
}
// check if all nodes are connected
for (int i = 0; i < n; i++) {
if (!visited[i])
return false;
}
return true;
}
}
Solution3_b Code：
class Solution {
private class NodeWithParent {
int id;
int parentId;
public NodeWithParent(int id, int parentId) {
this.id = id;
this.parentId = parentId;
}
}
public boolean validTree(int n, int[][] edges) {
// init graph (adjList)
List> adjList = new ArrayList>(n);
for (int i = 0; i < n; i++)
adjList.add(i, new ArrayList());
for (int i = 0; i < edges.length; i++) {
int u = edges[i][0], v = edges[i][1];
adjList.get(u).add(v);
adjList.get(v).add(u);
}
// dfs
boolean[] visited = new boolean[n];
Deque stack = new ArrayDeque();
stack.push(new NodeWithParent(0, -1));
visited[0] = true;
while (!stack.isEmpty()) {
NodeWithParent cur = stack.pop();
visited[cur.id] = true;
for (int neighborId: adjList.get(cur.id)) {
if (neighborId == cur.parentId) {
continue;
}
else if ((visited[neighborId])) {
return false;
}
else {
stack.push(new NodeWithParent(neighborId, cur.id));
}
}
}
// check if all nodes are connected
for (int i = 0; i < n; i++) {
if (!visited[i])
return false;
}
return true;
}
}
Solution4_a Code：
class Solution {
class UF {
private int[] id;
private int[] sz; // for an id, the number of elements in that id
private int count; // number of sort of id
public UF(int n) {
this.id = new int[n];
this.sz = new int[n];
this.count = 0;
// init
for (int i = 0; i < n; i++) {
this.id[i] = i;
this.sz[i] = 1;
this.count++;
}
}
public void union(int p, int q) {
int p_root = find(p), q_root = find(q);
// weighted quick union
///*
if(p_root == q_root) return;
if (sz[p_root] < sz[q_root]) {
id[p_root] = q_root; sz[q_root] += sz[p_root];
} else {
id[q_root] = p_root; sz[p_root] += sz[q_root];
}
--count;
//*/
// regular
/*
if(p_root == q_root) return;
id[p_root] = q_root;
--count;
*/
}
public int find(int i) { // path compression
for (;i != id[i]; i = id[i])
id[i] = id[id[i]];
return i;
}
public boolean connected(int p, int q) {
int p_root = find(p);
int q_root = find(q);
if(p_root != q_root) return false;
else return true;
}
public int count() {
return this.count;
}
}
public boolean validTree(int n, int[][] edges) {
UF uf = new UF(n);
// perform union find
for (int i = 0; i < edges.length; i++) {
int x = uf.find(edges[i][0]);
int y = uf.find(edges[i][1]);
// if two vertices happen to be in the same set
// then there's a cycle
if (x == y) return false;
else {
// union
uf.union(x, y);
}
}
return edges.length == n - 1;
}
}
Solution4_b Code：
public class Solution {
public boolean validTree(int n, int[][] edges) {
// initialize n isolated islands
int[] nums = new int[n];
Arrays.fill(nums, -1);
// perform union find
for (int i = 0; i < edges.length; i++) {
int x = find(nums, edges[i][0]);
int y = find(nums, edges[i][1]);
// if two vertices happen to be in the same set
// then there's a cycle
if (x == y) return false;
// union
nums[x] = y;
}
return edges.length == n - 1;
}
int find(int nums[], int i) {
if (nums[i] == -1) return i;
return find(nums, nums[i]);
}
}